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(B) According to the Stefan-Boltzmann law, the energy emitted per second (power) $P$ by a black body is given by $P = \sigma A T^4$, where $\sigma$ is

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$4$

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(B) According to the Stefan-Boltzmann law, the energy emitted per second (power) $P$ by a black body is given by $P = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the absolute temperature in Kelvin.Initial state: $P_1 = E$, $A_1 = A$, $T_1 = 27 + 273 = 300 \ K$.So, $E = \sigma A (300)^4$.Final state: Length and breadth are halved, so $A_2 = (L/2) 	imes (B/2) = A/4$. Temperature $T_2 = 327 + 273 = 600 \ K$.New power $P_2 = \sigma A_2 T_2^4 = \sigma (A/4) (600)^4$.Taking the ratio: $P_2 / E = [\sigma (A/4) (600)^4] / [\sigma A (300)^4] = (1/4) 	imes (600/300)^4 = (1/4) 	imes (2)^4 = (1/4) 	imes 16 = 4$.Therefore, $P_2 = 4E$.

$A$ black rectangular surface of area $A$ emits energy $E$ per second at $27^{\circ}C$. If the length and breadth are reduced to half of their initial values and the temperature is raised to $327^{\circ}C$, then the energy emitted per second becomes: (in $E$)

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